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How does a 555 timer in monostable mode produce a single timed pulse when triggered?

The 555 monostable: producing a single output pulse when triggered, the pulse-duration equation, and using a monostable for timed delays and switch debouncing.

An Eduqas GCSE Electronics answer on the 555 timer in monostable mode: how a trigger produces a single output pulse, the pulse-duration equation, choosing the timing resistor and capacitor for a target time, and using a monostable for timed delays and switch debouncing.

Generated by Claude Opus 4.812 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

Eduqas wants you to use the 555 timer in monostable mode: how a trigger produces a single output pulse, the pulse-duration equation, choosing the timing resistor and capacitor for a target time, and using a monostable for timed delays and switch debouncing. The monostable is the standard way to produce one fixed-length pulse on demand.

The answer

The monostable and its single pulse

The pulse-duration equation

Choosing the components

Timed delays and debouncing

Examples in context

The monostable is the standard timed-pulse block: porch and stairwell lights, camera flash timing, and any "press to start a fixed delay" function are monostables driving a transistor switch. Switch debouncing with a monostable cleans up the input to counters and microcontrollers, linking this topic to the sequential module. The same RC product that sets the pulse here sets the astable frequency and the time constant, so all the timing calculations share one core idea.

Try this

Q1. State the equation for the output pulse duration of a 555 monostable. [1 mark]

  • Cue. T=1.1 RCT = 1.1\,RC.

Q2. A monostable has R=47 kΩR = 47\ \text{k}\Omega and C=10 μFC = 10\ \mu\text{F}. Find the pulse duration. [2 marks]

  • Cue. T=1.1×47000×10×10−6=0.52 sT = 1.1 \times 47000 \times 10 \times 10^{-6} = 0.52\ \text{s}.

Q3. State one use of a 555 monostable. [1 mark]

  • Cue. A timed delay (or switch debouncing, or stretching a short input into a fixed-length pulse).

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20204 marksA 555 monostable must produce an output pulse of 2.0 s2.0\ \text{s} when triggered, using a 100 μF100\ \mu\text{F} capacitor. Calculate the timing resistor required.
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Use the monostable pulse-duration equation T=1.1 RCT = 1.1\,RC, rearranged for the resistor: R=T1.1 CR = \dfrac{T}{1.1\,C}.

Substitute (capacitance in farads): R=2.01.1×100×10−6=2.01.1×10−4=2.01.1×104=1.8×104 Ω=18 kΩR = \dfrac{2.0}{1.1 \times 100 \times 10^{-6}} = \dfrac{2.0}{1.1 \times 10^{-4}} = \dfrac{2.0}{1.1} \times 10^{4} = 1.8 \times 10^{4}\ \Omega = 18\ \text{k}\Omega.

Markers reward the relation T=1.1RCT = 1.1 RC, the rearrangement, the capacitance in farads, and the resistor of about 18 kΩ18\ \text{k}\Omega.

Eduqas 20224 marksExplain the difference between a monostable and an astable 555 circuit, and give one use of a monostable.
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Difference (up to 3 marks): a monostable has one stable state (output low) and produces a single output pulse of fixed length each time it is triggered; it then returns to its stable state and waits. An astable has no stable state and free-runs, producing a continuous square wave with no trigger needed.

Use (1 mark): a timed delay (for example a light that stays on for a set time after a button is pressed), or debouncing a switch, or stretching a short input into a fixed-length pulse.

Markers reward the single-triggered-pulse versus continuous distinction and a valid monostable use.

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